ba=(16/(16+8))bd=(16/24)bd=(2/3)bd area of the rectangle = bd*bf area of triangle = (2/3)bd*bf/2=(1/3)*bd*bf 1/3
Because it can be derived from a large number of more basic principles applied to numbers. Since a/b and c/d are defined, then b and d are non-zero. Then a/b = ad/bd since multiplying the numerator and denominator of a fraction by a non-zero number leaves it unchanged. Similarly, c/d = bc/bd [These steps implicitly assume the commutative property of multiplication ie there is no difference between premultiplication and postmultiplication] Another property of numbers: If x = p and y = q then x+y = p+q [equals added to equals result in equals]. So a/b + c+d = ad/bd + bc/bd and finally, using the distributive property of multiplication [by 1/bd] over addition, = (ad + bc)/bd Maybe you wish you hadn't asked!
Yes, for example (a + bi)(c + di) = ac + adi + bic + bidi, and commutative property works as well --> ac + adi + bci + bdi² --> ac + (ad + bc)i + bd(-1) = (ac - bd) + (ad + bc)i
A tetrahedron ABCD is isosceles if AB=BD, AC=BD, and AD=BC. In other words, in an isosceles tetrahedron the opposite edges are equal. See the related link for more information.
(a + b)/(a - b) = (c + d)/(c - d) cross multiply(a + b)(c - d) = (a - b)(c + d)ac - ad + bc - bd = ac + ad - bc - bd-ad + bc = -bc + ad-ad - ad = - bc - bc-2ad = -2bcad = bc that is the product of the means equals the product of the extremesa/b = b/c
If bd ≠ 0, then a/b + c/d (the common denominator is bd) = (a x d)/(b x d) + (c x b)/(d x b) = ad/bd + cb/db = ad/bd + cb/bd = (ad + cb)/ bd
AD means that Anno Domini , the year of lord and BD means that Before Christ
Infinitely many ways!Suppose you have the fraction 2/d.Pick any pair of integers a and b where b � 0.Then 2b-ad is and integer, as is bd so that (2b - ad)/bd is a fraction.Consider the fractions a/b and (2b - ad)/bdTheir sum isa/b + (2b-ad)/bd = ad/bd + (2b-ad)/bd = 2b/bd = 2/d - as required.Since a and b were chosen arbitrarily, there are infinitely many possible answers to the question.
Yes. Since they are rational numbers, let's call the first one a/b and the second one c/d where a,b,c, and d are integers. Now we can subtract by finding a common denominator. Let's use bd. So we have ad/bd-cb/bc= (ad-bc)/CD which is rational since we know ad and bc are integers being the product of integers and CD is also an integers. Call ad-bd=P and call CD=Q where P and Q are integers. We now see the difference is of two rationals is rational.
A rational number is one that can be expressed as a/b The sum of two rational numbers is: a/b + c/d =ad/bd + bc/bd =(ad+bc)/bd =e/f which is rational The difference of two rational numbers is: a/b - c/d =ab/bd - bc/bd =(ab-bc)/bd =e/f which is rational The product of two rational numbers is: (a/b)(c/d) =ac/bd =e/f which is rational
The difference of two rational numbers is rational. Let the two rational numbers be a/b and c/d, where a, b, c, and d are integers. Any rational number can be represented this way. Their difference is a/b-c/d = ad/bd-cb/bd = (ad-cb)/bd. Products and differences of integers are always integers. This means that ad-cb is an integer, and so is bd. Thus, (ad-cb)/bd is a rational number (since it is the ratio of two integers). This is equivalent to the difference of the original two rational numbers.
ba=(16/(16+8))bd=(16/24)bd=(2/3)bd area of the rectangle = bd*bf area of triangle = (2/3)bd*bf/2=(1/3)*bd*bf 1/3
Really bad bd bd bd bd bd bd bd bd i dont know bd bd bd bd bd bd bd bd bd bd bd bd bd bd bd bd bd
It is 16 units.
Because it can be derived from a large number of more basic principles applied to numbers. Since a/b and c/d are defined, then b and d are non-zero. Then a/b = ad/bd since multiplying the numerator and denominator of a fraction by a non-zero number leaves it unchanged. Similarly, c/d = bc/bd [These steps implicitly assume the commutative property of multiplication ie there is no difference between premultiplication and postmultiplication] Another property of numbers: If x = p and y = q then x+y = p+q [equals added to equals result in equals]. So a/b + c+d = ad/bd + bc/bd and finally, using the distributive property of multiplication [by 1/bd] over addition, = (ad + bc)/bd Maybe you wish you hadn't asked!
its AD and BC, and AD means anno domini (greek) means after the birth of christ and BC stands for before christ,before the birth of jesus. RG
according to sum of fractions: a/b + c/d = (ad+bc)/bd hence for example: a=c=1 , b=d=2: (ad+bc)/bd = (2+2)/2x2 = 4/4 = 1 sagy